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gregori [183]
3 years ago
6

The mass-energy equivalence formula is an example of the law of the conservation of _________.

Physics
1 answer:
pishuonlain [190]3 years ago
6 0
D.matter and energy .................
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Suppose you are standing on the center of a merry-go-round that is at rest. You are holding a spinning bicycle wheel over your h
ioda

Answer:

B. It begins to rotate counterclockwise

Explanation:

It can happen that the merry-go-round remains at rest in the case when the person is standing on the axle  of the merry go round whose axis is fixed to some rigid support, but here the person is standing at the center of the merry-go-round not at the axle hence the according to the conservation of angular momentum. Angular momentum is given as:

L=I.\omega

for the conservation of momentum

I.\omega=I'.\omega'

where:

I= moment of inertia of the merry-go-round

\omega= angular velocity of the merry-go-round

I'= moment of inertia of the bicycle wheel

\omega'= angular velocity of the wheel

As the person tries to stop the wheel which is rotating then the person feels an opposing force which tries to moves the whole system in its direction.

6 0
3 years ago
Why do electrical devices have resistance​
Kamila [148]
As electrons move through the conductor, some collide with atoms, other electrons, or impurities in the metal.
8 0
3 years ago
A motorboat travels 92 km in 2 hours going upstream. It travels 132 km going downstream in the same amount of time. What is the
taurus [48]

Answer:

The speed on boat in still water is  56 \frac{km}{h}  and the rate of the current is   10 \frac{km}{h}

Explanation:

Since speed , v= \frac{Distance\, traveled(D)}{Time\, taken(t)}

Therefore speed of motor boat while traveling upstream is

v_{upstream}=\frac{92}{2}\frac{km}{h}=46\frac{km}{h}

and  speed of motor boat while traveling downstream is

v_{downstream}=\frac{132}{2}\frac{km}{h}=66\frac{km}{h}

Let speed of boat in still water be v_b and rate of current be v_w

Therefore v_{upstream}=v_b-v_w=46\frac{km}{h}   ----(A)

and  v_{downstream}=v_b+v_w=66\frac{km}{h}     ------(B)

Adding equation (A) and (B)  we get

2v_b= (46+66) \frac{km}{h}=112 \frac{km}{h}

=>v_b= 56 \frac{km}{h}   ------(C)

Substituting the value of  v_b in equation (A) we get

v_w= 10 \frac{km}{h}

Thus the speed on boat in still water is  56 \frac{km}{h}  and the rate of the current is   10 \frac{km}{h}

5 0
4 years ago
The length of a vector is its_____
zlopas [31]

Answer:

The answer here would be its "Magnitude".

5 0
4 years ago
Read 2 more answers
The normal force of a parked car is 15,000 Newtons. The coefficient of static friction between the rubber of the tires and the a
Shtirlitz [24]

Answer:

11250 N

Explanation:

From the question given above, the following data were obtained:

Normal force (R) = 15000 N

Coefficient of static friction (μ) = 0.75

Frictional force (F) =?

Friction and normal force are related by the following equation:

F = μR

Where:

F is the frictional force.

μ is the coefficient of static friction.

R is the normal force.

With the above formula, we can calculate the frictional force acting on the car as follow:

Normal force (R) = 15000 N

Coefficient of static friction (μ) = 0.75

Frictional force (F) =?

F = μR

F = 0.75 × 15000

F = 11250 N

Therefore, the frictional force acting on the car is 11250 N

3 0
3 years ago
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